## digitalmars.D - modulus and array.length

```Hi,

I've noticed that the modulus operator acts differently when the divisor is
the length of an array and the dividend is negative.
For instance, this code:

import std.stdio;
const int x = 4;
void main(){
int[x] arr = [0, 1, 2, 3];

writeln("Using arr.length");
for(int i = -3; i <= 0; i++){
writefln("%d mod %d = %d", i, arr.length, i % arr.length);
}
writeln("Using x");
for(int i = -3; i <= 0; i++){
writefln("%d mod %d = %d", i, x, i % x);
}
}

Produces this output:

Using arr.length
-3 mod 4 = 1
-2 mod 4 = 2
-1 mod 4 = 3
0 mod 4 = 0
Using x
-3 mod 4 = -3
-2 mod 4 = -2
-1 mod 4 = -1
0 mod 4 = 0

Which is a useful (but undocumented?) feature that lets you loop through
arrays backwards.
However, when the length of the array is odd...
const int x = 3;
...
int[x] arr = [0, 1, 2];
It looks like this:

Using arr.length
-3 mod 3 = 1  <-- this should be 0
-2 mod 3 = 2  <-- this should be 1
-1 mod 3 = 0  <-- this should be 2
0 mod 3 = 0
Using x
-3 mod 3 = 0
-2 mod 3 = -2
-1 mod 3 = -1
0 mod 3 = 0

Does anyone know of a reason for this? It doesn't seem like a bug, but I
don't know why it would do something like that.

Upon further investigation, I found that when arr.length is even, i acts
like (x*(abs(i)/x + 1) + i) , and when arr.length is odd, i acts like
(x*(abs(i)/x - 1) + i)

(I'm using dmd 2.049 on linux)

~Dave
```
Nov 16 2010
```On Tue, 16 Nov 2010 10:55:24 -0500, David Osborne <krendilboove gmail.com>
wrote:

Hi,

I've noticed that the modulus operator acts differently when the divisor
is
the length of an array and the dividend is negative.
For instance, this code:

import std.stdio;
const int x = 4;
void main(){
int[x] arr = [0, 1, 2, 3];

writeln("Using arr.length");
for(int i = -3; i <= 0; i++){
writefln("%d mod %d = %d", i, arr.length, i % arr.length);
}
writeln("Using x");
for(int i = -3; i <= 0; i++){
writefln("%d mod %d = %d", i, x, i % x);
}
}

Produces this output:

Using arr.length
-3 mod 4 = 1
-2 mod 4 = 2
-1 mod 4 = 3
0 mod 4 = 0
Using x
-3 mod 4 = -3
-2 mod 4 = -2
-1 mod 4 = -1
0 mod 4 = 0

Which is a useful (but undocumented?) feature that lets you loop through
arrays backwards.
However, when the length of the array is odd...
const int x = 3;
...
int[x] arr = [0, 1, 2];
It looks like this:

Using arr.length
-3 mod 3 = 1  <-- this should be 0
-2 mod 3 = 2  <-- this should be 1
-1 mod 3 = 0  <-- this should be 2
0 mod 3 = 0
Using x
-3 mod 3 = 0
-2 mod 3 = -2
-1 mod 3 = -1
0 mod 3 = 0

Does anyone know of a reason for this? It doesn't seem like a bug, but I
don't know why it would do something like that.

Upon further investigation, I found that when arr.length is even, i acts
like (x*(abs(i)/x + 1) + i) , and when arr.length is odd, i acts like
(x*(abs(i)/x - 1) + i)

(I'm using dmd 2.049 on linux)

Simple, array.length is an unsigned value, so the signed operands are
promoted to unsigned through normal integer promotion:

cast(uint)-3 -> 4_294_967_293u
4_294_967_293u % 3u -> 1

I don't think this "undocumented" feature is valid.  In fact, it may not
be valid for evens as well that aren't powers of 2.

-Steve
```
Nov 16 2010
```David Osborne wrote:
Using arr.length
-3 mod 3 =3D 1  <-- this should be 0
-2 mod 3 =3D 2  <-- this should be 1
-1 mod 3 =3D 0  <-- this should be 2
0 mod 3 =3D 0

Like others have said, this is due to the fact that when one
operand is unsigned (here the length), then all operands get casted
to unsigned before the operation. Several people have asked that
array length be changed to a signed value because of that
(bearophile probably has a bug report for it ;) )

Using x
-3 mod 3 =3D 0
-2 mod 3 =3D -2
-1 mod 3 =3D -1
0 mod 3 =3D 0
=20

This is consistent with the C behaviour. Note that according to the
C standard, the result of the modulus operation with negative
arguments is undefined. I don't remember if Walter has specified the
expected behaviour for D or if he left it undefined...

Note that from a strictly mathematical point of view, this result
is valid: for all x and all n, x-(x%n) is a multiple of n.

Jerome
--=20
mailto:jeberger free.fr
http://jeberger.free.fr
Jabber: jeberger jabber.fr
```
Nov 16 2010
```Jérôme M. Berger Wrote:

-3 mod 3 = 0
-2 mod 3 = -2
-1 mod 3 = -1
0 mod 3 = 0

Note that from a strictly mathematical point of view, this result
is valid: for all x and all n, x-(x%n) is a multiple of n.

It's rather (x/n)+(x%n)==x
```
Nov 16 2010
```Kagamin wrote:
J=C3=A9r=C3=B4me M. Berger Wrote:
=20
-3 mod 3 =3D 0
-2 mod 3 =3D -2
-1 mod 3 =3D -1
0 mod 3 =3D 0

Note that from a strictly mathematical point of view, this result
is valid: for all x and all n, x-(x%n) is a multiple of n.

=20
It's rather (x/n)+(x%n)=3D=3Dx

That is (part of) the definition of the *remainder* operation. The
definition of *modulus* is looser.

Jerome
--=20
mailto:jeberger free.fr
http://jeberger.free.fr
Jabber: jeberger jabber.fr
```
Nov 16 2010
```J. M. Berger:

(bearophile probably has a bug report for it ;) )

Yup, bug 3843.

Bye,
bearophile
```
Nov 16 2010
```Thanks for the info, everyone.  I guess I'll just have to be more careful
what I use modulus with :)

~Dave
```
Nov 16 2010